The restriction on the
domain comes from the fact that I can't divide by zero, soxcan't be equal to–2.
I usually wouldn't bother writing down the restriction, but it's helpful
here because I need to know the domain and range of the inverse. Note
from the picture (and recalling the concept of horizontal
asymptotes) thatywill never equal1.
Then the domain is "xis not equal to
–2" and the
range is " yis not equal to
1". For the
inverse, they'll be swapped: the domain will be "xis not equal to
1"and the range
will be "yis not equal to
–2". Here's
the algebra:

The
original function:

I
rename "f(x)"
as "y":

Then
I solve for "x=":

I
get thex-stuff
on one side:

Here's
the trick: I factor out thex!

Then
I switchxandy:

And
rename "y"
as "f-inverse";
the domain restriction comes from the fact that this is a
rational function.

Since the inverse
is just a rational function, then the inverse is indeed a function.

Here's the graph:

Thenthe
inverse is y
= (–2x – 2) / (x
– 1),
and the inverse is also a function, with domain of all xnot equal
to
1and
range of all ynot equal
to
–2.

Find the inverse
of f(x)
= x2 – 3x + 2, x< 1.5

With the domain
restriction, the graph looks like this:

From what I know
about graphing
quadratics,
the vertex is at (x,
y) = (1.5, –0.25),
so this graph is the left-hand "half" of the parabola.